If `E=cos^(2)71^(@)+cos^(2)49^(@)+cos71^(@) cos 49^(@)`, then the value of 10E is equal to

Options

A. 7.5
B. 2.5
C. 3.5
D. 4.5

Answer

To solve the problem \( E = \cos^2 71^\circ + \cos^2 49^\circ + \cos 71^\circ \cos 49^\circ \), we will break it down step by step.

### Step 1: Rewrite the expression for \( E \)
We start with the expression:
\[ E = \cos^2 71^\circ + \cos^2 49^\circ + \cos 71^\circ \cos 49^\circ \]

### Step 2: Use the identity for the sum of squares
Recall the identity:
\[ \cos^2 A + \cos^2 B + 2 \cos A \cos B = (\cos A + \cos B)^2 \]
Thus, we can rewrite \( E \) as:
\[ E = \cos^2 71^\circ + \cos^2 49^\circ + 2 \cos 71^\circ \cos 49^\circ - \cos 71^\circ \cos 49^\circ \]
This can be simplified to:
\[ E = \left( \cos 71^\circ + \cos 49^\circ \right)^2 - \cos 71^\circ \cos 49^\circ \]

### Step 3: Calculate \( \cos 71^\circ + \cos 49^\circ \)
Using the formula for the sum of cosines:
\[ \cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) \]
Substituting \( A = 71^\circ \) and \( B = 49^\circ \):
\[ \cos 71^\circ + \cos 49^\circ = 2 \cos\left(\frac{71^\circ + 49^\circ}{2}\right) \cos\left(\frac{71^\circ - 49^\circ}{2}\right) \]
Calculating the angles:
\[ \frac{71^\circ + 49^\circ}{2} = \frac{120^\circ}{2} = 60^\circ \]
\[ \frac{71^\circ - 49^\circ}{2} = \frac{22^\circ}{2} = 11^\circ \]
Thus,
\[ \cos 71^\circ + \cos 49^\circ = 2 \cos 60^\circ \cos 11^\circ = 2 \cdot \frac{1}{2} \cdot \cos 11^\circ = \cos 11^\circ \]

### Step 4: Substitute back into \( E \)
Now substituting back:
\[ E = (\cos 11^\circ)^2 - \cos 71^\circ \cos 49^\circ \]

### Step 5: Calculate \( \cos 71^\circ \cos 49^\circ \)
Using the product-to-sum formula:
\[ \cos A \cos B = \frac{1}{2} \left( \cos(A + B) + \cos(A - B) \right) \]
Substituting \( A = 71^\circ \) and \( B = 49^\circ \):
\[ \cos 71^\circ \cos 49^\circ = \frac{1}{2} \left( \cos(120^\circ) + \cos(22^\circ) \right) \]
Since \( \cos 120^\circ = -\frac{1}{2} \):
\[ \cos 71^\circ \cos 49^\circ = \frac{1}{2} \left( -\frac{1}{2} + \cos 22^\circ \right) = \frac{\cos 22^\circ - \frac{1}{2}}{2} \]

### Step 6: Final expression for \( E \)
Now substituting this back into \( E \):
\[ E = \cos^2 11^\circ - \frac{\cos 22^\circ - \frac{1}{2}}{2} \]

### Step 7: Calculate \( E \)
Using the identity \( \cos^2 A = \frac{1 + \cos 2A}{2} \):
\[ \cos^2 11^\circ = \frac{1 + \cos 22^\circ}{2} \]
Thus,
\[ E = \frac{1 + \cos 22^\circ}{2} - \frac{\cos 22^\circ - \frac{1}{2}}{2} \]
Simplifying this:
\[ E = \frac{1 + \cos 22^\circ - \cos 22^\circ + \frac{1}{2}}{2} = \frac{1 + \frac{1}{2}}{2} = \frac{3/2}{2} = \frac{3}{4} \]

### Step 8: Calculate \( 10E \)
Finally, we calculate:
\[ 10E = 10 \times \frac{3}{4} = \frac{30}{4} = 7.5 \]

Thus, the value of \( 10E \) is \( \boxed{7.5} \).